今天在某教材上看到一个二阶方阵的求逆公式 想推导一下
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If M=[abcd], then M−1=1det⁡M[d−b−ca]\text{If}\space \mathbf{M}= \begin{bmatrix} a & b\\ c & d \end{bmatrix} \text{, then}\space \mathbf{M}^{-1}=\frac{1}{\det \mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}

Proof 1.14.514\bold{Proof}\space\bold{1.14.514}
Let\text{Let}

M=[abcd]\mathbf{M}= \begin{bmatrix} a & b\\ c & d \end{bmatrix}

we want to find M−1 so that\text{we want to find}\space\mathbf{M}^{-1}\space\text{so that}

MM−1=I\mathbf{M}\mathbf{M}^{-1}=\mathbf{I}

Let\text{Let}

M[xy]=[x0y0](*)\mathbf{M} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} x_0\\ y_0 \end{bmatrix} \tag{*}

which is equivalent as\text{which is equivalent as}

{ax+by=x0cx+dy=y0\begin{cases} ax+by=x_0\\ cx+dy=y_0 \end{cases}

solved as\text{solved as}

{x=dad−bcx0+−bad−bcy0y=−cad−bcx0+aad−bcy0\begin{cases} x=\frac{d}{ad-bc}x_0+\frac{-b}{ad-bc}y_0\\ y=\frac{-c}{ad-bc}x_0+\frac{a}{ad-bc}y_0 \end{cases}

so we have\text{so we have}

[xy]=[dad−bc−bad−bc−cad−bcaad−bc][x0y0]=1ad−bc[d−b−ca][x0y0]=1det⁡M[d−b−ca][x0y0]\begin{aligned} \begin{bmatrix} x\\ y \end{bmatrix} &= \begin{bmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix}\\ &= \frac{1}{ad-bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix}\\ &= \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix} \end{aligned}

back to (∗), we have\text{back to}\space(*)\text{, we have}

M1det⁡M[d−b−ca][x0y0]=[x0y0]\mathbf{M} \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix} = \begin{bmatrix} x_0\\ y_0 \end{bmatrix}

so\text{so}

M1det⁡M[d−b−ca]=I\mathbf{M} \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} = \mathbf{I}

here comes\text{here comes}

M−1=1det⁡M[d−b−ca]\mathbf{M}^{-1}= \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}

Q.E.D.\text{Q.E.D.}